## Tag `035E`

## 28.48. Normalization

In this section we construct the

normalization, and thenormalization of one scheme in another.Lemma 28.48.1. Let $X$ be a scheme. Let $\mathcal{A}$ be a quasi-coherent sheaf of $\mathcal{O}_X$-algebras. The subsheaf $\mathcal{A}' \subset \mathcal{A}$ defined by the rule $$ U \longmapsto \{f \in \mathcal{A}(U) \mid f_x \in \mathcal{A}_x \text{ integral over } \mathcal{O}_{X, x} \text{ for all }x \in U\} $$ is a quasi-coherent $\mathcal{O}_X$-algebra, the stalk $\mathcal{A}'_x$ is the integral closure of $\mathcal{O}_{X, x}$ in $\mathcal{A}_x$, and for any affine open $U \subset X$ the ring $\mathcal{A}'(U) \subset \mathcal{A}(U)$ is the integral closure of $\mathcal{O}_X(U)$ in $\mathcal{A}(U)$.

Proof.This is a subsheaf by the local nature of the conditions. It is an $\mathcal{O}_X$-algebra by Algebra, Lemma 10.35.7. Let $U \subset X$ be an affine open. Say $U = \mathop{\rm Spec}(R)$ and say $\mathcal{A}$ is the quasi-coherent sheaf associated to the $R$-algebra $A$. Then according to Algebra, Lemma 10.35.10 the value of $\mathcal{A}'$ over $U$ is given by the integral closure $A'$ of $R$ in $A$. This proves the last assertion of the lemma. To prove that $\mathcal{A}'$ is quasi-coherent, it suffices to show that $\mathcal{A}'(D(f)) = A'_f$. This follows from the fact that integral closure and localization commute, see Algebra, Lemma 10.35.9. The same fact shows that the stalks are as advertised. $\square$Definition 28.48.2. Let $X$ be a scheme. Let $\mathcal{A}$ be a quasi-coherent sheaf of $\mathcal{O}_X$-algebras. The

integral closure of $\mathcal{O}_X$ in $\mathcal{A}$is the quasi-coherent $\mathcal{O}_X$-subalgebra $\mathcal{A}' \subset \mathcal{A}$ constructed in Lemma 28.48.1 above.In the setting of the definition above we can consider the morphism of relative spectra $$ \xymatrix{ Y = \underline{\mathop{\rm Spec}}_X(\mathcal{A}) \ar[rr] \ar[rd] & & X' = \underline{\mathop{\rm Spec}}_X(\mathcal{A}') \ar[ld] \\ & X & } $$ see Lemma 28.13.5. The scheme $X' \to X$ will be the normalization of $X$ in the scheme $Y$. Here is a slightly more general setting. Suppose we have a quasi-compact and quasi-separated morphism $f : Y \to X$ of schemes. In this case the sheaf of $\mathcal{O}_X$-algebras $f_*\mathcal{O}_Y$ is quasi-coherent, see Schemes, Lemma 25.24.1. Taking the integral closure $\mathcal{O}' \subset f_*\mathcal{O}_Y$ we obtain a quasi-coherent sheaf of $\mathcal{O}_X$-algebras whose relative spectrum is the normalization of $X$ in $Y$. Here is the formal definition.

Definition 28.48.3. Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of schemes. Let $\mathcal{O}'$ be the integral closure of $\mathcal{O}_X$ in $f_*\mathcal{O}_Y$. The

normalization of $X$ in $Y$is the scheme^{1}$$ \nu : X' = \underline{\mathop{\rm Spec}}_X(\mathcal{O}') \to X $$ over $X$. It comes equipped with a natural factorization $$ Y \xrightarrow{f'} X' \xrightarrow{\nu} X $$ of the initial morphism $f$.The factorization is the composition of the canonical morphism $Y \to \underline{\mathop{\rm Spec}}(f_*\mathcal{O}_Y)$ (see Constructions, Lemma 26.4.7) and the morphism of relative spectra coming from the inclusion map $\mathcal{O}' \to f_*\mathcal{O}_Y$. We can characterize the normalization as follows.

Lemma 28.48.4. Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of schemes. The factorization $f = \nu \circ f'$, where $\nu : X' \to X$ is the normalization of $X$ in $Y$ is characterized by the following two properties:

- the morphism $\nu$ is integral, and
- for any factorization $f = \pi \circ g$, with $\pi : Z \to X$ integral, there exists a commutative diagram $$ \xymatrix{ Y \ar[d]_{f'} \ar[r]_g & Z \ar[d]^\pi \\ X' \ar[ru]^h \ar[r]^\nu & X } $$ for some unique morphism $h : X' \to Z$.
Moreover, in (2) the morphism $h : X' \to Z$ is the normalization of $Z$ in $Y$.

Proof.Let $\mathcal{O}' \subset f_*\mathcal{O}_Y$ be the integral closure of $\mathcal{O}_X$ as in Definition 28.48.3. The morphism $\nu$ is integral by construction, which proves (1). Assume given a factorization $f = \pi \circ g$ with $\pi : Z \to X$ integral as in (2). By Definition 28.44.1 $\pi$ is affine, and hence $Z$ is the relative spectrum of a quasi-coherent sheaf of $\mathcal{O}_X$-algebras $\mathcal{B}$. The morphism $g : X \to Z$ corresponds to a map of $\mathcal{O}_X$-algebras $\chi : \mathcal{B} \to f_*\mathcal{O}_Y$. Since $\mathcal{B}(U)$ is integral over $\mathcal{O}_X(U)$ for every affine open $U \subset X$ (by Definition 28.44.1) we see from Lemma 28.48.1 that $\chi(\mathcal{B}) \subset \mathcal{O}'$. By the functoriality of the relative spectrum Lemma 28.13.5 this provides us with a unique morphism $h : X' \to Z$. We omit the verification that the diagram commutes.It is clear that (1) and (2) characterize the factorization $f = \nu \circ f'$ since it characterizes it as an initial object in a category. The morphism $h$ in (2) is integral by Lemma 28.44.12. Given a factorization $g = \pi' \circ g'$ with $\pi' : Z' \to Z$ integral, we get a factorization $f = (\pi \circ \pi') \circ g'$ and we get a morphism $h' : X' \to Z'$. Uniqueness implies that $\pi' \circ h' = h$. Hence the characterization (1), (2) applies to the morphism $h : X' \to Z$ which gives the last statement of the lemma. $\square$

Lemma 28.48.5. Let $$ \xymatrix{ Y_2 \ar[d]_{f_2} \ar[r] & Y_1 \ar[d]^{f_1} \\ X_2 \ar[r] & X_1 } $$ be a commutative diagram of morphisms of schemes. Assume $f_1$, $f_2$ quasi-compact and quasi-separated. Let $f_i = \nu_i \circ f_i'$, $i = 1, 2$ be the canonical factorizations, where $\nu_i : X_i' \to X_i$ is the normalization of $X_i$ in $Y_i$. Then there exists a canonical commutative diagram $$ \xymatrix{ Y_2 \ar[d]_{f_2'} \ar[r] & Y_1 \ar[d]^{f_1'} \\ X_2' \ar[d]_{\nu_2} \ar[r] & X_1' \ar[d]^{\nu_1} \\ X_2 \ar[r] & X_1 } $$

Proof.By Lemmas 28.48.4 (1) and 28.44.6 the base change $X_2 \times_{X_1} X'_1 \to X_2$ is integral. Note that $f_2$ factors through this morphism. Hence we get a canonical morphism $X'_2 \to X_2 \times_{X_1} X'_1$ from Lemma 28.48.4 (2). This gives the middle horizontal arrow in the last diagram. $\square$Lemma 28.48.6. Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of schemes. Let $U \subset X$ be an open subscheme and set $V = f^{-1}(U)$. Then the normalization of $U$ in $V$ is the inverse image of $U$ in the normalization of $X$ in $Y$.

Proof.Clear from the construction. $\square$Lemma 28.48.7. Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of schemes. Let $X' \to X$ be the normalization of $X$ in $Y$. If $Y$ is reduced, so is $X'$.

Proof.This follows from the fact that a subring of a reduced ring is reduced. Some details omitted. $\square$Lemma 28.48.8. Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of schemes. Let $X' \to X$ be the normalization of $X$ in $Y$. Every generic point of an irreducible component of $X'$ is the image of a generic point of an irreducible component of $Y$.

Proof.By Lemma 28.48.6 we may assume $X = \mathop{\rm Spec}(A)$ is affine. Choose a finite affine open covering $Y = \bigcup \mathop{\rm Spec}(B_i)$. Then $X' = \mathop{\rm Spec}(A')$ and the morphisms $\mathop{\rm Spec}(B_i) \to Y \to X'$ jointly define an injective $A$-algebra map $A' \to \prod B_i$. Thus the lemma follows from Algebra, Lemma 10.29.5. $\square$Lemma 28.48.9. Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of schemes. Suppose that $Y = Y_1 \amalg Y_2$ is a disjoint union of two schemes. Write $f_i = f|_{Y_i}$. Let $X_i'$ be the normalization of $X$ in $Y_i$. Then $X_1' \amalg X_2'$ is the normalization of $X$ in $Y$.

Proof.In terms of integral closures this corresponds to the following fact: Let $A \to B$ be a ring map. Suppose that $B = B_1 \times B_2$. Let $A_i'$ be the integral closure of $A$ in $B_i$. Then $A_1' \times A_2'$ is the integral closure of $A$ in $B$. The reason this works is that the elements $(1, 0)$ and $(0, 1)$ of $B$ are idempotents and hence integral over $A$. Thus the integral closure $A'$ of $A$ in $B$ is a product and it is not hard to see that the factors are the integral closures $A'_i$ as described above (some details omitted). $\square$Lemma 28.48.10. Let $f : X \to S$ be a quasi-compact, quasi-separated and universally closed morphisms of schemes. Then $f_*\mathcal{O}_X$ is integral over $\mathcal{O}_S$. In other words, the normalization of $S$ in $X$ is equal to the factorization $$ X \longrightarrow \underline{\mathop{\rm Spec}}_S(f_*\mathcal{O}_X) \longrightarrow S $$ of Constructions, Lemma 26.4.7.

Proof.The question is local on $S$, hence we may assume $S = \mathop{\rm Spec}(R)$ is affine. Let $h \in \Gamma(X, \mathcal{O}_X)$. We have to show that $h$ satisfies a monic equation over $R$. Think of $h$ as a morphism as in the following commutative diagram $$ \xymatrix{ X \ar[rr]_h \ar[rd]_f & & \mathbf{A}^1_S \ar[ld] \\ & S & } $$ Let $Z \subset \mathbf{A}^1_S$ be the scheme theoretic image of $h$, see Definition 28.6.2. The morphism $h$ is quasi-compact as $f$ is quasi-compact and $\mathbf{A}^1_S \to S$ is separated, see Schemes, Lemma 25.21.15. By Lemma 28.6.3 the morphism $X \to Z$ is dominant. By Lemma 28.42.7 the morphism $X \to Z$ is closed. Hence $h(X) = Z$ (set theoretically). Thus we can use Lemma 28.42.8 to conclude that $Z \to S$ is universally closed (and even proper). Since $Z \subset \mathbf{A}^1_S$, we see that $Z \to S$ is affine and proper, hence integral by Lemma 28.44.7. Writing $\mathbf{A}^1_S = \mathop{\rm Spec}(R[T])$ we conclude that the ideal $I \subset R[T]$ of $Z$ contains a monic polynomial $P(T) \in R[T]$. Hence $P(h) = 0$ and we win. $\square$Lemma 28.48.11. Let $f : Y \to X$ be an integral morphism. Then the integral closure of $X$ in $Y$ is equal to $Y$.

Proof.By Lemma 28.44.7 this is a special case of Lemma 28.48.10. $\square$Lemma 28.48.12. Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of schemes. Assume

- $Y$ is a normal scheme,
- quasi-compact opens of $Y$ have finitely many irreducible components.
Then the normalization $X'$ of $X$ in $Y$ is a normal scheme. Moreover, the morphism $Y \to X'$ is dominant and induces a bijection of irreducible components.

Proof.We first prove that $X'$ is normal. Let $U \subset X$ be an affine open. It suffices to prove that the inverse image of $U$ in $X'$ is normal (see Properties, Lemma 27.7.2). By Lemma 28.48.6 we may replace $X$ by $U$, and hence we may assume $X = \mathop{\rm Spec}(A)$ affine. In this case $Y$ is quasi-compact, and hence has a finite number of irreducible components by assumption. Hence $Y = \coprod_{i = 1, \ldots n} Y_i$ is a finite disjoint union of normal integral schemes by Properties, Lemma 27.7.5. By Lemma 28.48.9 we see that $X' = \coprod_{i = 1, \ldots, n} X_i'$, where $X'_i$ is the normalization of $X$ in $Y_i$. By Properties, Lemma 27.7.9 we see that $B_i = \Gamma(Y_i, \mathcal{O}_{Y_i})$ is a normal domain. Note that $X_i' = \mathop{\rm Spec}(A_i')$, where $A_i' \subset B_i$ is the integral closure of $A$ in $B_i$, see Lemma 28.48.1. By Algebra, Lemma 10.36.2 we see that $A_i' \subset B_i$ is a normal domain. Hence $X' = \coprod X_i'$ is a finite union of normal schemes and hence is normal.It is clear from the description of $X'$ above that $Y \to X'$ is dominant and induces a bijection on irreducible components if $X$ is affine. The result in general follows from this by a topological argument (omitted). $\square$

Lemma 28.48.13. Let $f : X \to S$ be a morphism. Assume that

- $S$ is a Nagata scheme,
- $f$ is quasi-compact and quasi-separated,
- quasi-compact opens of $X$ have finitely many irreducible components,
- if $x \in X$ is a generic point of an irreducible component, then the field extension $\kappa(f(x)) \subset \kappa(x)$ is finitely generated, and
- $X$ is reduced.
Then the normalization $\nu : S' \to S$ of $S$ in $X$ is finite.

Proof.There is an immediate reduction to the case $S = \mathop{\rm Spec}(R)$ where $R$ is a Nagata ring by assumption (1). We have to show that the integral closure $A$ of $R$ in $\Gamma(X, \mathcal{O}_X)$ is finite over $R$. Since $f$ is quasi-compact by assumption (2) we can write $X = \bigcup_{i = 1, \ldots, n} U_i$ with each $U_i$ affine. Say $U_i = \mathop{\rm Spec}(B_i)$. Each $B_i$ is reduced by assumption (5) and has finitely many minial primes $\mathfrak q_{i1}, \ldots, \mathfrak q_{im_i}$ by assumption (3) and Algebra, Lemma 10.25.1. We have $$ \Gamma(X, \mathcal{O}_X) \subset B_1 \times \ldots \times B_n \subset \prod\nolimits_{i = 1, \ldots, n} \prod\nolimits_{j = 1, \ldots, m_i} (B_i)_{\mathfrak q_{ij}} $$ the second inclusion by Algebra, Lemma 10.24.2. We have $\kappa(\mathfrak q_{ij}) = (B_i)_{\mathfrak q_{ij}}$ by Algebra, Lemma 10.24.1. Hence the integral closure $A$ of $R$ in $\Gamma(X, \mathcal{O}_X)$ is contained in the product of the integral closures $A_{ij}$ of $R$ in $\kappa(\mathfrak q_{ij})$. Since $R$ is Noetherian it suffices to show that $A_{ij}$ is a finite $R$-module for each $i, j$. Let $\mathfrak p_{ij} \subset R$ be the image of $\mathfrak q_{ij}$. As $\kappa(\mathfrak p_{ij}) \subset \kappa(\mathfrak q_{ij})$ is a finitely generated field extension by assumption (4), we see that $R \to \kappa(\mathfrak q_{ij})$ is essentially of finite type. Thus $R \to A_{ij}$ is finite by Algebra, Lemma 10.151.17. $\square$Lemma 28.48.14. Let $f : X \to S$ be a morphism. Assume that

- $S$ is a Nagata scheme,
- $f$ is of finite type,
- $X$ is reduced.
Then the normalization $\nu : S' \to S$ of $S$ in $X$ is finite.

Proof.This is a special case of Lemma 28.48.13. Namely, (2) holds as the finite type morphism $f$ is quasi-compact by definition and quasi-separated by Lemma 28.16.7. Condition (3) holds because $X$ is locally Noetherian by Lemma 28.16.6. Finally, condition (4) holds because a finite type morphism induces finitely generated residue field extensions. $\square$Next, we come to the normalization of a scheme $X$. We only define/construct it when $X$ has locally finitely many irreducible components. Let $X$ be a scheme such that every quasi-compact open has finitely many irreducible components. Let $X^{(0)} \subset X$ be the set of generic points of irreducible components of $X$. Let \begin{equation} \tag{28.48.14.1} f : Y = \coprod\nolimits_{\eta \in X^{(0)}} \mathop{\rm Spec}(\kappa(\eta)) \longrightarrow X \end{equation} be the inclusion of the generic points into $X$ using the canonical maps of Schemes, Section 25.13. Note that this morphism is quasi-compact by assumption and quasi-separated as $Y$ is separated (see Schemes, Section 25.21).

Definition 28.48.15. Let $X$ be a scheme such that every quasi-compact open has finitely many irreducible components. We define the

normalizationof $X$ as the morphism $$ \nu : X^\nu \longrightarrow X $$ which is the normalization of $X$ in the morphism $f : Y \to X$ (28.48.14.1) constructed above.Any locally Noetherian scheme has a locally finite set of irreducible components and the definition applies to it. Usually the normalization is defined only for reduced schemes. With the definition above the normalization of $X$ is the same as the normalization of the reduction $X_{red}$ of $X$.

Lemma 28.48.16. Let $X$ be a scheme such that every quasi-compact open has finitely many irreducible components. The normalization morphism $\nu$ factors through the reduction $X_{red}$ and $X^\nu \to X_{red}$ is the normalization of $X_{red}$.

Proof.Let $f : Y \to X$ be the morphism (28.48.14.1). We get a factorization $Y \to X_{red} \to X$ of $f$ from Schemes, Lemma 25.12.6. By Lemma 28.48.4 we obtain a canonical morphism $X^\nu \to X_{red}$ and that $X^\nu$ is the normalization of $X_{red}$ in $Y$. The lemma follows as $Y \to X_{red}$ is identical to the morphism (28.48.14.1) constructed for $X_{red}$. $\square$If $X$ is reduced, then the normalization of $X$ is the same as the relative spectrum of the integral closure of $\mathcal{O}_X$ in the sheaf of meromorphic functions $\mathcal{K}_X$ (see Divisors, Section 30.17). Namely, $\mathcal{K}_X = f_*\mathcal{O}_Y$ in this case, see Divisors, Lemma 30.17.8 and its proof. We describe this here explicitly.

Lemma 28.48.17. Let $X$ be a reduced scheme such that every quasi-compact open has finitely many irreducible components. Let $\mathop{\rm Spec}(A) = U \subset X$ be an affine open. Then

- $A$ has finitely many minimal primes $\mathfrak q_1, \ldots, \mathfrak q_t$,
- the total ring of fractions $Q(A)$ of $A$ is $Q(A/\mathfrak q_1) \times \ldots \times Q(A/\mathfrak q_t)$,
- the integral closure $A'$ of $A$ in $Q(A)$ is the product of the integral closures of the domains $A/\mathfrak q_i$ in the fields $Q(A/\mathfrak q_i)$, and
- $\nu^{-1}(U)$ is identified with the spectrum of $A'$.

Proof.Minimal primes correspond to irreducible components (Algebra, Lemma 10.25.1), hence we have (1) by assumption. Then $(0) = \mathfrak q_1 \cap \ldots \cap \mathfrak q_t$ because $A$ is reduced (Algebra, Lemma 10.16.2). Then we have $Q(A) = \prod A_{\mathfrak q_i} = \prod \kappa(\mathfrak q_i)$ by Algebra, Lemmas 10.24.4 and 10.24.1. This proves (2). Part (3) follows from Algebra, Lemma 10.36.14, or Lemma 28.48.9. Part (4) holds because it is clear that $f^{-1}(U) \to U$ is the morphism $$ \mathop{\rm Spec}\left(\prod \kappa(\mathfrak q_i)\right) \longrightarrow \mathop{\rm Spec}(A) $$ where $f : Y \to X$ is the morphism (28.48.14.1). $\square$Lemma 28.48.18. Let $X$ be a scheme such that every quasi-compact open has finitely many irreducible components.

- The normalization $X^\nu$ is a normal scheme.
- The morphism $\nu : X^\nu \to X$ is integral, surjective, and induces a bijection on irreducible components.
- For any integral, birational
^{2}morphism $X' \to X$ there exists a factorization $X^\nu \to X' \to X$ and $X^\nu \to X'$ is the normalization of $X'$.- For any morphism $Z \to X$ with $Z$ a normal scheme such that each irreducible component of $Z$ dominates an irreducible component of $X$ there exists a unique factorization $Z \to X^\nu \to X$.

Proof.Let $f : Y \to X$ be as in (28.48.14.1). Part (1) follows from Lemma 28.48.12 and the fact that $Y$ is normal. It also follows from the description of the affine opens in Lemma 28.48.17.The morphism $\nu$ is integral by Lemma 28.48.4. By Lemma 28.48.12 the morphism $Y \to X^\nu$ induces a bijection on irreducible components, and by construction of $Y$ this implies that $X^\nu \to X$ induces a bijection on irreducible components. By construction $f : Y \to X$ is dominant, hence also $\nu$ is dominant. Since an integral morphism is closed (Lemma 28.44.7) this implies that $\nu$ is surjective. This proves (2).

Suppose that $\alpha : X' \to X$ is integral and birational. Any quasi-compact open $U'$ of $X'$ maps to a quasi-compact open of $X$, hence we see that $U'$ has only finitely many irreducible components. Let $f' : Y' \to X'$ be the morphism (28.48.14.1) constructed starting with $X'$. As $\alpha$ is birational it is clear that $Y' = Y$ and $f = \alpha \circ f'$. Hence the factorization $X^\nu \to X' \to X$ exists and $X^\nu \to X'$ is the normalization of $X'$ by Lemma 28.48.4. This proves (3).

Let $g : Z \to X$ be a morphism whose domain is a normal scheme and such that every irreducible component dominates an irreducible component of $X$. By Lemma 28.48.16 we have $X^\nu = X_{red}^\nu$ and by Schemes, Lemma 25.12.6 $Z \to X$ factors through $X_{red}$. Hence we may replace $X$ by $X_{red}$ and assume $X$ is reduced. Moreover, as the factorization is unique it suffices to construct it locally on $Z$. Let $W \subset Z$ and $U \subset X$ be affine opens such that $g(W) \subset U$. Write $U = \mathop{\rm Spec}(A)$ and $W = \mathop{\rm Spec}(B)$, with $g|_W$ given by $\varphi : A \to B$. We will use the results of Lemma 28.48.17 freely. Let $\mathfrak p_1, \ldots, \mathfrak p_t$ be the minimal primes of $A$. As $Z$ is normal, we see that $B$ is a normal ring, in particular reduced. Moreover, by assumption any minimal prime $\mathfrak q \subset B$ we have that $\varphi^{-1}(\mathfrak q)$ is a minimal prime of $A$. Hence if $x \in A$ is a nonzerodivisor, i.e., $x \not \in \bigcup \mathfrak p_i$, then $\varphi(x)$ is a nonzerodivisor in $B$. Thus we obtain a canonical ring map $Q(A) \to Q(B)$. As $B$ is normal it is equal to its integral closure in $Q(B)$ (see Algebra, Lemma 10.36.11). Hence we see that the integral closure $A' \subset Q(A)$ of $A$ maps into $B$ via the canonical map $Q(A) \to Q(B)$. Since $\nu^{-1}(U) = \mathop{\rm Spec}(A')$ this gives the canonical factorization $W \to \nu^{-1}(U) \to U$ of $\nu|_W$. We omit the verification that it is unique. $\square$

Lemma 28.48.19. A finite (or even integral) birational morphism $f : X \to Y$ of integral schemes with $Y$ normal is an isomorphism.

Proof.Let $V \subset Y$ be an affine open with inverse image $U \subset X$ which is an affine open too. Since $f$ is a birational morphism of integral schemes, the homomorphism $\mathcal{O}_Y(V) \to \mathcal{O}_X(U)$ is an injective map of domains which induces an isomorphism of fraction fields. As $Y$ is normal, the ring $\mathcal{O}_Y(V)$ is integrally closed in the fraction field. Since $f$ is finite (or integral) every element of $\mathcal{O}_X(U)$ is integral over $\mathcal{O}_Y(V)$. We conclude that $\mathcal{O}_Y(V) = \mathcal{O}_X(U)$. This proves that $f$ is an isomorphism as desired. $\square$Lemma 28.48.20. Let $X$ be an integral, Japanese scheme. The normalization $\nu : X^\nu \to X$ is a finite morphism.

Proof.Follows from the definitions and Lemma 28.48.17. Namely, in this case the lemma says that $\nu^{-1}(\mathop{\rm Spec}(A))$ is the spectrum of the integral closure of $A$ in its field of fractions. $\square$Lemma 28.48.21. Let $X$ be a Nagata scheme. The normalization $\nu : X^\nu \to X$ is a finite morphism.

Proof.Note that a Nagata scheme is locally Noetherian, thus Definition 28.48.15 does apply. The lemma is now a special case of Lemma 28.48.13 but we can also prove it directly as follows. Write $X^\nu \to X$ as the composition $X^\nu \to X_{red} \to X$. As $X_{red} \to X$ is a closed immersion it is finite. Hence it suffices to prove the lemma for a reduced Nagata scheme (by Lemma 28.44.5). Let $\mathop{\rm Spec}(A) = U \subset X$ be an affine open. By Lemma 28.48.17 we have $\nu^{-1}(U) = \mathop{\rm Spec}(\prod A_i')$ where $A_i'$ is the integral closure of $A/\mathfrak q_i$ in its fraction field. As $A$ is a Nagata ring (see Properties, Lemma 27.13.6) each of the ring extensions $A/\mathfrak q_i \subset A'_i$ are finite. Hence $A \to \prod A'_i$ is a finite ring map and we win. $\square$

The code snippet corresponding to this tag is a part of the file `morphisms.tex` and is located in lines 10784–11447 (see updates for more information).

```
\section{Normalization}
\label{section-normalization}
\noindent
In this section we construct the {\it normalization}, and the
{\it normalization of one scheme in another}.
\begin{lemma}
\label{lemma-integral-closure}
Let $X$ be a scheme. Let $\mathcal{A}$ be a quasi-coherent sheaf
of $\mathcal{O}_X$-algebras. The subsheaf $\mathcal{A}' \subset \mathcal{A}$
defined by the rule
$$
U \longmapsto \{f \in \mathcal{A}(U) \mid
f_x \in \mathcal{A}_x \text{ integral over } \mathcal{O}_{X, x}
\text{ for all }x \in U\}
$$
is a quasi-coherent $\mathcal{O}_X$-algebra, the stalk $\mathcal{A}'_x$
is the integral closure of $\mathcal{O}_{X, x}$ in $\mathcal{A}_x$, and
for any affine open $U \subset X$ the ring
$\mathcal{A}'(U) \subset \mathcal{A}(U)$ is
the integral closure of $\mathcal{O}_X(U)$ in $\mathcal{A}(U)$.
\end{lemma}
\begin{proof}
This is a subsheaf by the local nature of the conditions.
It is an $\mathcal{O}_X$-algebra by
Algebra, Lemma \ref{algebra-lemma-integral-closure-is-ring}.
Let $U \subset X$ be an affine open. Say $U = \Spec(R)$
and say $\mathcal{A}$ is the quasi-coherent sheaf associated to
the $R$-algebra $A$. Then according to
Algebra, Lemma \ref{algebra-lemma-integral-closure-stalks}
the value of $\mathcal{A}'$ over $U$ is given by the integral
closure $A'$ of $R$ in $A$. This proves the last assertion of
the lemma. To prove that $\mathcal{A}'$ is quasi-coherent, it
suffices to show that $\mathcal{A}'(D(f)) = A'_f$. This follows
from the fact that integral closure and localization commute, see
Algebra, Lemma \ref{algebra-lemma-integral-closure-localize}.
The same fact shows that the stalks are as advertised.
\end{proof}
\begin{definition}
\label{definition-integral-closure}
Let $X$ be a scheme. Let $\mathcal{A}$ be a quasi-coherent sheaf
of $\mathcal{O}_X$-algebras. The {\it integral closure of $\mathcal{O}_X$
in $\mathcal{A}$} is the quasi-coherent $\mathcal{O}_X$-subalgebra
$\mathcal{A}' \subset \mathcal{A}$ constructed in
Lemma \ref{lemma-integral-closure} above.
\end{definition}
\noindent
In the setting of the definition above we can consider the morphism
of relative spectra
$$
\xymatrix{
Y = \underline{\Spec}_X(\mathcal{A}) \ar[rr] \ar[rd] & &
X' = \underline{\Spec}_X(\mathcal{A}') \ar[ld] \\
& X &
}
$$
see Lemma \ref{lemma-affine-equivalence-algebras}.
The scheme $X' \to X$ will be the normalization of $X$ in the scheme $Y$.
Here is a slightly more general setting. Suppose we have a
quasi-compact and quasi-separated morphism $f : Y \to X$
of schemes. In this case the sheaf of
$\mathcal{O}_X$-algebras $f_*\mathcal{O}_Y$ is quasi-coherent, see
Schemes, Lemma \ref{schemes-lemma-push-forward-quasi-coherent}.
Taking the integral closure $\mathcal{O}' \subset f_*\mathcal{O}_Y$
we obtain a quasi-coherent sheaf of $\mathcal{O}_X$-algebras
whose relative spectrum is the normalization of $X$ in $Y$. Here is
the formal definition.
\begin{definition}
\label{definition-normalization-X-in-Y}
Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of schemes.
Let $\mathcal{O}'$ be the integral closure of $\mathcal{O}_X$ in
$f_*\mathcal{O}_Y$. The {\it normalization of $X$ in $Y$} is the
scheme\footnote{The scheme $X'$ need not be normal, for example if
$Y = X$ and $f = \text{id}_X$, then $X' = X$.}
$$
\nu : X' = \underline{\Spec}_X(\mathcal{O}') \to X
$$
over $X$. It comes equipped with a natural factorization
$$
Y \xrightarrow{f'} X' \xrightarrow{\nu} X
$$
of the initial morphism $f$.
\end{definition}
\noindent
The factorization is the composition of the canonical morphism
$Y \to \underline{\Spec}(f_*\mathcal{O}_Y)$ (see
Constructions, Lemma
\ref{constructions-lemma-canonical-morphism})
and the morphism of relative spectra coming from the inclusion map
$\mathcal{O}' \to f_*\mathcal{O}_Y$. We can characterize the
normalization as follows.
\begin{lemma}
\label{lemma-characterize-normalization}
Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of schemes.
The factorization $f = \nu \circ f'$, where $\nu : X' \to X$ is the
normalization of $X$ in $Y$ is characterized by the following
two properties:
\begin{enumerate}
\item the morphism $\nu$ is integral, and
\item for any factorization $f = \pi \circ g$, with $\pi : Z \to X$
integral, there exists a commutative diagram
$$
\xymatrix{
Y \ar[d]_{f'} \ar[r]_g & Z \ar[d]^\pi \\
X' \ar[ru]^h \ar[r]^\nu & X
}
$$
for some unique morphism $h : X' \to Z$.
\end{enumerate}
Moreover, in (2) the morphism $h : X' \to Z$ is the normalization of
$Z$ in $Y$.
\end{lemma}
\begin{proof}
Let $\mathcal{O}' \subset f_*\mathcal{O}_Y$ be the integral closure of
$\mathcal{O}_X$ as in Definition \ref{definition-normalization-X-in-Y}.
The morphism $\nu$ is integral by construction, which proves (1).
Assume given a factorization $f = \pi \circ g$ with $\pi : Z \to X$
integral as in (2). By Definition \ref{definition-integral}
$\pi$ is affine, and hence $Z$ is the relative
spectrum of a quasi-coherent sheaf of $\mathcal{O}_X$-algebras $\mathcal{B}$.
The morphism $g : X \to Z$ corresponds to a map of $\mathcal{O}_X$-algebras
$\chi : \mathcal{B} \to f_*\mathcal{O}_Y$. Since $\mathcal{B}(U)$ is
integral over $\mathcal{O}_X(U)$ for every affine open $U \subset X$
(by Definition \ref{definition-integral})
we see from Lemma \ref{lemma-integral-closure}
that $\chi(\mathcal{B}) \subset \mathcal{O}'$.
By the functoriality of the relative spectrum
Lemma \ref{lemma-affine-equivalence-algebras}
this provides us with a unique morphism
$h : X' \to Z$. We omit the verification that the diagram commutes.
\medskip\noindent
It is clear that (1) and (2) characterize the
factorization $f = \nu \circ f'$ since it characterizes it
as an initial object in a category. The morphism $h$ in (2)
is integral by Lemma \ref{lemma-finite-permanence}.
Given a factorization $g = \pi' \circ g'$ with $\pi' : Z' \to Z$
integral, we get a factorization $f = (\pi \circ \pi') \circ g'$ and
we get a morphism $h' : X' \to Z'$. Uniqueness implies that
$\pi' \circ h' = h$. Hence the characterization (1), (2) applies
to the morphism $h : X' \to Z$ which gives the last statement of the lemma.
\end{proof}
\begin{lemma}
\label{lemma-functoriality-normalization}
Let
$$
\xymatrix{
Y_2 \ar[d]_{f_2} \ar[r] & Y_1 \ar[d]^{f_1} \\
X_2 \ar[r] & X_1
}
$$
be a commutative diagram of morphisms of schemes.
Assume $f_1$, $f_2$ quasi-compact and quasi-separated.
Let $f_i = \nu_i \circ f_i'$, $i = 1, 2$
be the canonical factorizations, where $\nu_i : X_i' \to X_i$ is
the normalization of $X_i$ in $Y_i$. Then there exists a
canonical commutative diagram
$$
\xymatrix{
Y_2 \ar[d]_{f_2'} \ar[r] & Y_1 \ar[d]^{f_1'} \\
X_2' \ar[d]_{\nu_2} \ar[r] & X_1' \ar[d]^{\nu_1} \\
X_2 \ar[r] & X_1
}
$$
\end{lemma}
\begin{proof}
By Lemmas \ref{lemma-characterize-normalization} (1)
and \ref{lemma-base-change-finite}
the base change $X_2 \times_{X_1} X'_1 \to X_2$
is integral. Note that $f_2$ factors through this morphism.
Hence we get a canonical morphism
$X'_2 \to X_2 \times_{X_1} X'_1$ from
Lemma \ref{lemma-characterize-normalization} (2).
This gives the middle horizontal arrow in the last diagram.
\end{proof}
\begin{lemma}
\label{lemma-normalization-localization}
Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of schemes.
Let $U \subset X$ be an open subscheme and set $V = f^{-1}(U)$.
Then the normalization of $U$ in $V$ is the inverse image of $U$
in the normalization of $X$ in $Y$.
\end{lemma}
\begin{proof}
Clear from the construction.
\end{proof}
\begin{lemma}
\label{lemma-normalization-in-reduced}
Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of schemes.
Let $X' \to X$ be the normalization of $X$ in $Y$. If $Y$ is reduced, so
is $X'$.
\end{lemma}
\begin{proof}
This follows from the fact that a subring of a reduced ring is reduced.
Some details omitted.
\end{proof}
\begin{lemma}
\label{lemma-normalization-generic}
Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of schemes.
Let $X' \to X$ be the normalization of $X$ in $Y$. Every generic point of
an irreducible component of $X'$ is the image of a generic point of
an irreducible component of $Y$.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-normalization-localization} we may assume $X = \Spec(A)$
is affine. Choose a finite affine open covering $Y = \bigcup \Spec(B_i)$.
Then $X' = \Spec(A')$ and the morphisms $\Spec(B_i) \to Y \to X'$
jointly define an injective $A$-algebra map $A' \to \prod B_i$.
Thus the lemma follows from
Algebra, Lemma \ref{algebra-lemma-injective-minimal-primes-in-image}.
\end{proof}
\begin{lemma}
\label{lemma-normalization-in-disjoint-union}
Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of schemes.
Suppose that $Y = Y_1 \amalg Y_2$ is a disjoint union of two schemes.
Write $f_i = f|_{Y_i}$. Let $X_i'$ be the normalization of $X$ in $Y_i$.
Then $X_1' \amalg X_2'$ is the normalization of $X$ in $Y$.
\end{lemma}
\begin{proof}
In terms of integral closures this corresponds to the following fact:
Let $A \to B$ be a ring map. Suppose that $B = B_1 \times B_2$.
Let $A_i'$ be the integral closure of $A$ in $B_i$. Then
$A_1' \times A_2'$ is the integral closure of $A$ in $B$.
The reason this works is that the elements $(1, 0)$ and $(0, 1)$ of $B$
are idempotents and hence integral over $A$. Thus the integral closure
$A'$ of $A$ in $B$ is a product and it is not hard to see that the factors
are the integral closures $A'_i$ as described above (some details
omitted).
\end{proof}
\begin{lemma}
\label{lemma-normalization-in-universally-closed}
Let $f : X \to S$ be a quasi-compact, quasi-separated and
universally closed morphisms of schemes.
Then $f_*\mathcal{O}_X$ is integral over $\mathcal{O}_S$. In other
words, the normalization of $S$ in $X$ is equal to the factorization
$$
X \longrightarrow \underline{\Spec}_S(f_*\mathcal{O}_X)
\longrightarrow S
$$
of Constructions, Lemma \ref{constructions-lemma-canonical-morphism}.
\end{lemma}
\begin{proof}
The question is local on $S$, hence we may assume $S = \Spec(R)$
is affine. Let $h \in \Gamma(X, \mathcal{O}_X)$. We have to show that
$h$ satisfies a monic equation over $R$. Think of $h$ as a morphism
as in the following commutative diagram
$$
\xymatrix{
X \ar[rr]_h \ar[rd]_f & & \mathbf{A}^1_S \ar[ld] \\
& S &
}
$$
Let $Z \subset \mathbf{A}^1_S$ be the scheme theoretic image of $h$,
see Definition \ref{definition-scheme-theoretic-image}.
The morphism $h$ is quasi-compact as $f$ is quasi-compact and
$\mathbf{A}^1_S \to S$ is separated, see
Schemes, Lemma \ref{schemes-lemma-quasi-compact-permanence}.
By Lemma \ref{lemma-quasi-compact-scheme-theoretic-image} the
morphism $X \to Z$ is dominant. By
Lemma \ref{lemma-image-proper-scheme-closed} the morphism
$X \to Z$ is closed. Hence $h(X) = Z$ (set theoretically).
Thus we can use
Lemma \ref{lemma-image-proper-is-proper}
to conclude that $Z \to S$ is universally closed (and even proper).
Since $Z \subset \mathbf{A}^1_S$, we see that $Z \to S$ is affine
and proper, hence integral by Lemma \ref{lemma-integral-universally-closed}.
Writing $\mathbf{A}^1_S = \Spec(R[T])$ we conclude that
the ideal $I \subset R[T]$ of $Z$ contains a monic polynomial
$P(T) \in R[T]$. Hence $P(h) = 0$ and we win.
\end{proof}
\begin{lemma}
\label{lemma-normalization-in-integral}
Let $f : Y \to X$ be an integral morphism.
Then the integral closure of $X$ in $Y$ is equal to $Y$.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-integral-universally-closed} this is a special case of
Lemma \ref{lemma-normalization-in-universally-closed}.
\end{proof}
\begin{lemma}
\label{lemma-normal-normalization}
Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism
of schemes. Assume
\begin{enumerate}
\item $Y$ is a normal scheme,
\item quasi-compact opens of $Y$ have finitely many irreducible components.
\end{enumerate}
Then the normalization $X'$ of $X$ in $Y$ is a normal scheme. Moreover,
the morphism $Y \to X'$ is dominant and induces a bijection of
irreducible components.
\end{lemma}
\begin{proof}
We first prove that $X'$ is normal.
Let $U \subset X$ be an affine open. It suffices to prove that
the inverse image of $U$ in $X'$ is normal (see
Properties, Lemma \ref{properties-lemma-locally-normal}).
By Lemma \ref{lemma-normalization-localization} we may replace $X$ by $U$,
and hence we may assume $X = \Spec(A)$ affine.
In this case $Y$ is quasi-compact, and hence has a finite number of
irreducible components by assumption. Hence
$Y = \coprod_{i = 1, \ldots n} Y_i$ is a finite disjoint union of
normal integral schemes by
Properties, Lemma \ref{properties-lemma-normal-locally-finite-nr-irreducibles}.
By Lemma \ref{lemma-normalization-in-disjoint-union}
we see that $X' = \coprod_{i = 1, \ldots, n} X_i'$,
where $X'_i$ is the normalization of $X$ in $Y_i$.
By Properties, Lemma \ref{properties-lemma-normal-integral-sections}
we see that $B_i = \Gamma(Y_i, \mathcal{O}_{Y_i})$ is a normal domain.
Note that $X_i' = \Spec(A_i')$, where $A_i' \subset B_i$
is the integral closure of $A$ in $B_i$, see
Lemma \ref{lemma-integral-closure}. By
Algebra, Lemma \ref{algebra-lemma-integral-closure-in-normal}
we see that $A_i' \subset B_i$ is a normal domain.
Hence $X' = \coprod X_i'$ is a finite union of normal schemes
and hence is normal.
\medskip\noindent
It is clear from the description of $X'$ above that $Y \to X'$
is dominant and induces a bijection on irreducible components if
$X$ is affine. The result in general follows from this by a topological
argument (omitted).
\end{proof}
\begin{lemma}
\label{lemma-nagata-normalization-finite-general}
Let $f : X \to S$ be a morphism. Assume that
\begin{enumerate}
\item $S$ is a Nagata scheme,
\item $f$ is quasi-compact and quasi-separated,
\item quasi-compact opens of $X$ have finitely many irreducible components,
\item if $x \in X$ is a generic point of an irreducible component,
then the field extension $\kappa(f(x)) \subset \kappa(x)$ is finitely
generated, and
\item $X$ is reduced.
\end{enumerate}
Then the normalization $\nu : S' \to S$ of $S$ in $X$ is finite.
\end{lemma}
\begin{proof}
There is an immediate reduction to the case $S = \Spec(R)$
where $R$ is a Nagata ring by assumption (1). We have to show that
the integral closure $A$ of $R$ in $\Gamma(X, \mathcal{O}_X)$ is
finite over $R$. Since $f$ is quasi-compact by assumption (2) we can write
$X = \bigcup_{i = 1, \ldots, n} U_i$ with each $U_i$ affine.
Say $U_i = \Spec(B_i)$. Each $B_i$ is reduced by assumption (5)
and has finitely many minial primes
$\mathfrak q_{i1}, \ldots, \mathfrak q_{im_i}$
by assumption (3) and
Algebra, Lemma \ref{algebra-lemma-irreducible}.
We have
$$
\Gamma(X, \mathcal{O}_X) \subset B_1 \times \ldots \times B_n
\subset
\prod\nolimits_{i = 1, \ldots, n}
\prod\nolimits_{j = 1, \ldots, m_i} (B_i)_{\mathfrak q_{ij}}
$$
the second inclusion by
Algebra, Lemma \ref{algebra-lemma-reduced-ring-sub-product-fields}.
We have $\kappa(\mathfrak q_{ij}) = (B_i)_{\mathfrak q_{ij}}$ by
Algebra, Lemma \ref{algebra-lemma-minimal-prime-reduced-ring}.
Hence the integral closure $A$ of $R$ in $\Gamma(X, \mathcal{O}_X)$
is contained in the product of the integral closures $A_{ij}$ of
$R$ in $\kappa(\mathfrak q_{ij})$. Since $R$ is Noetherian
it suffices to show that $A_{ij}$ is a finite $R$-module for each $i, j$.
Let $\mathfrak p_{ij} \subset R$ be the image of $\mathfrak q_{ij}$.
As $\kappa(\mathfrak p_{ij}) \subset \kappa(\mathfrak q_{ij})$
is a finitely generated field extension by assumption (4),
we see that $R \to \kappa(\mathfrak q_{ij})$ is essentially of finite type.
Thus $R \to A_{ij}$ is finite by Algebra, Lemma
\ref{algebra-lemma-nagata-in-reduced-finite-type-finite-integral-closure}.
\end{proof}
\begin{lemma}
\label{lemma-nagata-normalization-finite}
Let $f : X \to S$ be a morphism. Assume that
\begin{enumerate}
\item $S$ is a Nagata scheme,
\item $f$ is of finite type,
\item $X$ is reduced.
\end{enumerate}
Then the normalization $\nu : S' \to S$ of $S$ in $X$ is finite.
\end{lemma}
\begin{proof}
This is a special case of
Lemma \ref{lemma-nagata-normalization-finite-general}.
Namely, (2) holds as the finite type morphism $f$ is quasi-compact
by definition and quasi-separated by
Lemma \ref{lemma-finite-type-Noetherian-quasi-separated}.
Condition (3) holds because $X$ is locally Noetherian by
Lemma \ref{lemma-finite-type-noetherian}. Finally, condition (4)
holds because a finite type morphism induces finitely generated
residue field extensions.
\end{proof}
\noindent
Next, we come to the normalization of a scheme $X$.
We only define/construct it when $X$ has locally finitely many irreducible
components. Let $X$ be a scheme such that every quasi-compact open has
finitely many irreducible components. Let
$X^{(0)} \subset X$ be the set of generic points of irreducible components
of $X$. Let
\begin{equation}
\label{equation-generic-points}
f :
Y = \coprod\nolimits_{\eta \in X^{(0)}} \Spec(\kappa(\eta))
\longrightarrow
X
\end{equation}
be the inclusion of the generic points into $X$ using the
canonical maps of Schemes, Section \ref{schemes-section-points}.
Note that this morphism is quasi-compact by assumption and
quasi-separated as $Y$ is separated (see
Schemes, Section \ref{schemes-section-separation-axioms}).
\begin{definition}
\label{definition-normalization}
Let $X$ be a scheme such that every quasi-compact open has
finitely many irreducible components. We define the
{\it normalization} of $X$ as the morphism
$$
\nu : X^\nu \longrightarrow X
$$
which is the normalization of $X$ in the morphism $f : Y \to X$
(\ref{equation-generic-points}) constructed above.
\end{definition}
\noindent
Any locally Noetherian scheme has a locally finite set of irreducible
components and the definition applies to it.
Usually the normalization is defined only for reduced schemes.
With the definition above the normalization of $X$ is the same
as the normalization of the reduction $X_{red}$ of $X$.
\begin{lemma}
\label{lemma-normalization-reduced}
Let $X$ be a scheme such that every quasi-compact open has
finitely many irreducible components. The normalization morphism
$\nu$ factors through the reduction $X_{red}$ and $X^\nu \to X_{red}$
is the normalization of $X_{red}$.
\end{lemma}
\begin{proof}
Let $f : Y \to X$ be the morphism (\ref{equation-generic-points}).
We get a factorization $Y \to X_{red} \to X$ of $f$ from
Schemes, Lemma \ref{schemes-lemma-map-into-reduction}.
By Lemma \ref{lemma-characterize-normalization} we obtain a canonical
morphism $X^\nu \to X_{red}$
and that $X^\nu$ is the normalization of $X_{red}$ in $Y$.
The lemma follows as $Y \to X_{red}$ is identical to the morphism
(\ref{equation-generic-points}) constructed for $X_{red}$.
\end{proof}
\noindent
If $X$ is reduced, then the normalization of $X$ is the same
as the relative spectrum of the integral closure of $\mathcal{O}_X$
in the sheaf of meromorphic functions $\mathcal{K}_X$
(see Divisors, Section \ref{divisors-section-meromorphic-functions}).
Namely, $\mathcal{K}_X = f_*\mathcal{O}_Y$ in this case, see
Divisors, Lemma \ref{divisors-lemma-reduced-finite-irreducible}
and its proof. We describe this here explicitly.
\begin{lemma}
\label{lemma-description-normalization}
Let $X$ be a reduced scheme such that every quasi-compact open has
finitely many irreducible components. Let $\Spec(A) = U \subset X$
be an affine open. Then
\begin{enumerate}
\item $A$ has finitely many minimal primes
$\mathfrak q_1, \ldots, \mathfrak q_t$,
\item the total ring of fractions $Q(A)$ of $A$ is
$Q(A/\mathfrak q_1) \times \ldots \times Q(A/\mathfrak q_t)$,
\item the integral closure $A'$ of $A$ in $Q(A)$ is the product of
the integral closures of the domains $A/\mathfrak q_i$
in the fields $Q(A/\mathfrak q_i)$, and
\item $\nu^{-1}(U)$ is identified with the spectrum of $A'$.
\end{enumerate}
\end{lemma}
\begin{proof}
Minimal primes correspond to irreducible components
(Algebra, Lemma \ref{algebra-lemma-irreducible}),
hence we have (1) by assumption. Then
$(0) = \mathfrak q_1 \cap \ldots \cap \mathfrak q_t$ because $A$ is reduced
(Algebra, Lemma \ref{algebra-lemma-Zariski-topology}).
Then we have
$Q(A) = \prod A_{\mathfrak q_i} = \prod \kappa(\mathfrak q_i)$
by Algebra, Lemmas \ref{algebra-lemma-total-ring-fractions-no-embedded-points}
and \ref{algebra-lemma-minimal-prime-reduced-ring}.
This proves (2). Part (3) follows from
Algebra, Lemma \ref{algebra-lemma-characterize-reduced-ring-normal},
or Lemma \ref{lemma-normalization-in-disjoint-union}.
Part (4) holds because it is clear that $f^{-1}(U) \to U$ is the morphism
$$
\Spec\left(\prod \kappa(\mathfrak q_i)\right)
\longrightarrow
\Spec(A)
$$
where $f : Y \to X$ is the morphism (\ref{equation-generic-points}).
\end{proof}
\begin{lemma}
\label{lemma-normalization-normal}
Let $X$ be a scheme such that every quasi-compact open has
finitely many irreducible components.
\begin{enumerate}
\item The normalization $X^\nu$ is a normal scheme.
\item The morphism $\nu : X^\nu \to X$ is integral, surjective, and
induces a bijection on irreducible components.
\item For any integral, birational\footnote{It suffices if
$X'_{red} \to X_{red}$ is birational.} morphism $X' \to X$ there
exists a factorization $X^\nu \to X' \to X$ and $X^\nu \to X'$
is the normalization of $X'$.
\item For any morphism $Z \to X$ with $Z$ a normal scheme
such that each irreducible component of $Z$ dominates an irreducible
component of $X$ there exists a unique factorization $Z \to X^\nu \to X$.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $f : Y \to X$ be as in (\ref{equation-generic-points}).
Part (1) follows from Lemma \ref{lemma-normal-normalization}
and the fact that $Y$ is normal. It also follows from the description
of the affine opens in Lemma \ref{lemma-description-normalization}.
\medskip\noindent
The morphism $\nu$ is integral by Lemma \ref{lemma-characterize-normalization}.
By Lemma \ref{lemma-normal-normalization} the
morphism $Y \to X^\nu$ induces a bijection on irreducible components,
and by construction of $Y$ this implies that $X^\nu \to X$ induces
a bijection on irreducible components. By construction $f : Y \to X$
is dominant, hence also $\nu$ is dominant. Since an integral morphism is
closed (Lemma \ref{lemma-integral-universally-closed}) this implies that
$\nu$ is surjective. This proves (2).
\medskip\noindent
Suppose that $\alpha : X' \to X$ is integral and birational.
Any quasi-compact open $U'$ of $X'$ maps to a quasi-compact open
of $X$, hence we see that $U'$ has only finitely many irreducible
components. Let $f' : Y' \to X'$ be the morphism
(\ref{equation-generic-points}) constructed starting with $X'$.
As $\alpha$ is birational
it is clear that $Y' = Y$ and $f = \alpha \circ f'$. Hence
the factorization $X^\nu \to X' \to X$ exists
and $X^\nu \to X'$ is the normalization of $X'$ by
Lemma \ref{lemma-characterize-normalization}. This proves (3).
\medskip\noindent
Let $g : Z \to X$ be a morphism whose domain is a normal scheme
and such that every irreducible component dominates an irreducible
component of $X$. By Lemma \ref{lemma-normalization-reduced}
we have $X^\nu = X_{red}^\nu$ and by
Schemes, Lemma \ref{schemes-lemma-map-into-reduction}
$Z \to X$ factors through $X_{red}$. Hence we may replace $X$ by
$X_{red}$ and assume $X$ is reduced. Moreover, as the factorization
is unique it suffices to construct it locally on $Z$.
Let $W \subset Z$ and $U \subset X$ be affine opens
such that $g(W) \subset U$. Write $U = \Spec(A)$ and
$W = \Spec(B)$, with $g|_W$ given by $\varphi : A \to B$.
We will use the results of Lemma \ref{lemma-description-normalization} freely.
Let $\mathfrak p_1, \ldots, \mathfrak p_t$ be the minimal primes of $A$.
As $Z$ is normal, we see that $B$ is a normal
ring, in particular reduced. Moreover, by assumption any minimal
prime $\mathfrak q \subset B$ we have that $\varphi^{-1}(\mathfrak q)$
is a minimal prime of $A$. Hence if $x \in A$ is a nonzerodivisor, i.e.,
$x \not \in \bigcup \mathfrak p_i$, then $\varphi(x)$ is a nonzerodivisor
in $B$. Thus we obtain a canonical ring map $Q(A) \to Q(B)$. As $B$ is
normal it is equal to its integral closure in $Q(B)$ (see
Algebra, Lemma \ref{algebra-lemma-normal-ring-integrally-closed}).
Hence we see that the integral closure $A' \subset Q(A)$ of $A$
maps into $B$ via the canonical map $Q(A) \to Q(B)$.
Since $\nu^{-1}(U) = \Spec(A')$ this gives the canonical
factorization $W \to \nu^{-1}(U) \to U$ of $\nu|_W$.
We omit the verification that it is unique.
\end{proof}
\begin{lemma}
\label{lemma-finite-birational-over-normal}
A finite (or even integral) birational morphism $f : X \to Y$
of integral schemes with $Y$ normal is an isomorphism.
\end{lemma}
\begin{proof}
Let $V \subset Y$ be an affine open
with inverse image $U \subset X$ which is an affine open too.
Since $f$ is a birational morphism of integral schemes, the homomorphism
$\mathcal{O}_Y(V) \to \mathcal{O}_X(U)$ is an injective map of domains
which induces an isomorphism of fraction fields. As $Y$ is normal,
the ring $\mathcal{O}_Y(V)$ is integrally closed in the fraction field.
Since $f$ is finite (or integral) every element of $\mathcal{O}_X(U)$
is integral over $\mathcal{O}_Y(V)$. We conclude that
$\mathcal{O}_Y(V) = \mathcal{O}_X(U)$. This proves that $f$ is an
isomorphism as desired.
\end{proof}
\begin{lemma}
\label{lemma-Japanese-normalization}
Let $X$ be an integral, Japanese scheme.
The normalization $\nu : X^\nu \to X$ is a finite morphism.
\end{lemma}
\begin{proof}
Follows from the definitions and
Lemma \ref{lemma-description-normalization}. Namely, in this case
the lemma says that $\nu^{-1}(\Spec(A))$ is the spectrum
of the integral closure of $A$ in its field of fractions.
\end{proof}
\begin{lemma}
\label{lemma-nagata-normalization}
Let $X$ be a Nagata scheme.
The normalization $\nu : X^\nu \to X$ is a finite morphism.
\end{lemma}
\begin{proof}
Note that a Nagata scheme is locally Noetherian, thus
Definition \ref{definition-normalization}
does apply. The lemma is now a special case of
Lemma \ref{lemma-nagata-normalization-finite-general}
but we can also prove it directly as follows.
Write $X^\nu \to X$ as the composition
$X^\nu \to X_{red} \to X$. As $X_{red} \to X$ is a closed immersion
it is finite. Hence it suffices to prove the lemma for a reduced
Nagata scheme (by Lemma \ref{lemma-composition-finite}).
Let $\Spec(A) = U \subset X$ be an affine open.
By Lemma \ref{lemma-description-normalization} we have
$\nu^{-1}(U) = \Spec(\prod A_i')$ where $A_i'$ is the integral
closure of $A/\mathfrak q_i$ in its fraction field. As $A$ is a Nagata
ring (see Properties, Lemma \ref{properties-lemma-locally-nagata})
each of the ring extensions
$A/\mathfrak q_i \subset A'_i$ are finite. Hence $A \to \prod A'_i$
is a finite ring map and we win.
\end{proof}
```

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